Problem 112 - Chemistry

Question

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\) is 0.83 at \(375^{\circ} \mathrm{C} . \mathrm{A}\) 14.6-g sample of ammonia is placed in a \(4.00-\mathrm{L}\) flask and heated to \(375^{\circ} \mathrm{C}\). Calculate the concentrations of all the gases when equilibrium is reached.

Step-by-Step Solution

Answer

The solution involves setting up an ICE table, substituting the concentrations into the equilibrium expression and solving the resulting quadratic equation. Perform the appropriate calculations to find the equilibrium concentrations of NH3, N2, and H2.

Step 1: Interpret the Problem and Gather Data

The reaction is \(2 NH_{3}(g) \rightleftharpoons N_{2}(g) + 3 H_{2}(g)\) with equilibrium constant \(K_{c} = 0.83\). Start with a 14.6g sample of ammonia in a 4.00L. flask. To convert grams of NH₃ to moles, remember \(NH_{3}\)'s molar mass is about 17 g/mol.

Step 2: Set up ICE Table

An ICE table helps organize the change in concentrations throughout the reaction. In the table, I initial), C (change), and E (equilibrium) refer to the concentrations at these instances.\n\nInitial moles of ammonia = 14.6g / 17 g/mol = 0.859 moles. Thus, initial concentrations: [NH3] = 0.859 mol/4.00 L = 0.215 M, [N2] = [H2] = 0 M (none initially).\n\nThe changes in concentration will be: -2x for NH3 (as per the stoichiometric coefficient), +x for N2, and +3x for H2.\n\nThe equilibrium concentrations thus become: [NH3] = 0.215 - 2x M, [N2] = x M, and [H2] = 3x M. Here x is the change in concentration of N2 or 1/3 of the change in concentration of H3.

Step 3: Substitute the Values in the Expression for Kc

The expression for \(K_{c}\) is [N2][H2]^3 / [NH3]^2. Substitute the equilibrium concentrations into Kc expression and set it equal to 0.83: 0.83 = (x(3x)^3) / ((0.215-2x)^2). This simplifies into a quadratic equation.

Step 4: Solve the Quadratic Equation

Solving the quadratic equation will provide the value of x. Then substitute x back into the equilibrium concentrations to find the final values.\n\nSolving the quadratic equation will involve either factoring, completing the square, or using the quadratic formula.

Step 5: Check Your Answer

After solving, it's important to verify the answer by substituting the calculated values back into the \(K_{c}\) equation to verify that it gives the known \(K_{c}\) value of 0.83.

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \(K_{c}\), is a measure of the ratio of the concentration of products to reactants at equilibrium for a particular reaction at a given temperature. The equilibrium constant expression for a general reaction is determined from the balanced chemical equation. For the reaction \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g)\), the equilibrium constant expression is:

• \(K_{c} = \dfrac{[\mathrm{N}_{2}][\mathrm{H}_{2}]^3}{[\mathrm{NH}_{3}]^2}\)

In this expression, the concentrations of the gaseous reactants and products (in moles per liter) are raised to the power of their coefficients from the balanced equation. Understanding \(K_{c}\) is crucial as it helps in predicting the direction of the reaction, understanding reaction extent, and calculating concentrations at equilibrium. A value of \(K_{c} = 0.83\) at \(375^{\circ} \mathrm{C}\) signifies that at this temperature, the reaction does not fully proceed to products but reaches a balance between reactants and products. This means that neither the reactants nor the products are significantly favored.

ICE Table

An ICE table is a very handy tool used in chemistry to track the Initial concentrations, the Change in concentrations, and the Equilibrium concentrations of reactants and products in a chemical reaction. In the given problem, the use of an ICE table makes it easier to visualize and organize the changes that occur as the system reaches equilibrium. The process starts with the initial concentrations:

• [NH3] = 0.215 M (calculated from moles of ammonia in the flask volume).
• [N2] and [H2] both start at 0 M because they are initially absent.

As the reaction progresses towards equilibrium, we assign a variable \(x\) to represent the change in the concentration of \(N_2\). Based on the stoichiometric coefficients from the balanced equation:

• \(\text{Change in } [NH3] = -2x \)
• \(\text{Change in } [N2] = +x \)
• \(\text{Change in } [H2] = +3x \)

At equilibrium, the new concentrations are substituted back into the \(K_{c}\) expression. This methodology enables the systematic evaluation of how the reaction dynamics shift as the system stabilizes.

Quadratic Equation

Solving a quadratic equation is often necessary in chemistry when dealing with equilibrium problems, especially when changes in concentrations are represented by variables. In this particular problem, after substituting the equilibrium concentrations into the \(K_{c}\) expression, we obtained:\[0.83 = \frac{x(3x)^3}{(0.215 - 2x)^2}\]This equation simplifies to a form that can be solved using the quadratic formula. A quadratic equation is typically written as \(ax^2 + bx + c = 0\). In this case, the expression becomes more manageable when expanded and simplified.To solve the quadratic equation:

• First, express the equation in standard form, identifying coefficients \(a\), \(b\), and \(c\).
• Use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Solving for \(x\) gives the changes in concentration, which are then used to find the equilibrium concentrations of all species. Remember to check that the solution is realistic given the initial assumptions, such as no negative concentrations, to ensure a valid chemical interpretation.